# BS 7671 Voltage Drop Calculation

## Contents

clc;
clear;
format compact;


## Example Cable Calculation BS 7671 Voltage Drop

For calculation details please refer to https://mycableengineering.com/knowledge-base/bs-7671-voltage-drop.

## Cable Details

table = '4E3A Single-core armoured 90 °C themosetting, copper conductors';

CSA = 50;       % cross sectional area of cable, mm2
Length = 80;    % length of cable, m
%
Uo = 400;       % line-line voltage, V
Ib = 152;       % load current, A
pf = 0.87;      % power factor
phases = 3;     % number of phases (0 = d.c.)
tp = 90;        % maximum permitted cable temperature, °C


Ensure that we have both cos(ph) and sin(phi) from the power factor.

cos_phi = pf;
sin_phi = sin(acos(pf));
disp('');
disp(['pf = cos(phi) = ', num2str(cos_phi), ' which gives sin(phi)= ', num2str(sin_phi,  '%0.2f')]);

pf = cos(phi) = 0.87 which gives sin(phi)= 0.49


## BS 7671 Looup Values

It = 222;       % tablulated current A, table 4E3A, column 6
mVr = 0.86;     % (mV/A/m)r table 4E3B, column 5
mVx = 0.29;     % (mV/A/m)x table 4E3B, column 5
Ca = 0.96;      % ambient temperature derating, table 4B1
Cg = 1.00;      % grouping derating, [no grouping]
Cs = 1.00;      % soil thermal resistivyt derating [not buired]
Cd = 1.00;      % depth of burial derating [not buried]


## Calculation

Ct = (230+tp-(Ca^2*Cg^2*Cs^2*Cd^2-(Ib^2/It^2))*(tp-30))/(230+tp);
disp(['Temperature correction factor Ct = ', num2str(Ct, '%0.2f' )]);

vd = ( Ct*cos_phi*mVr + sin_phi*mVx ) * Ib * Length /1000;
vdPercent = 100*vd/Uo;

disp('Voltage drop = ', num2str(vd, '%0.2f'), ' V [', num2str(vdPercent, '%0.2f'), '%]');

Temperature correction factor Ct = 0.92
Voltage drop = 10.06 V [2.52%] david chen Feb 19, 2018 8:28 PM Steven McFadyen Feb 19, 2018 10:33 PM